U.P.B. Sci. Bull., Series A, Vol. 77, Iss. 2, 2015 ISSN 1223-7027

A CLASSIFICATION OF CUBIC EDGE-TRANSITIVE GRAPHS OF ORDER 18p

Mehdi ALAEIYAN1, M.K. HOSSEINIPOOR2

A graph is called edge-transitive if its automorphism group acts transitively on the set of its edge. In this paper, we classify all connected cubic edge-transitive graphs of order 18p for each prime p.

Keywords: Edge-transitive graphs, Semisymmetric graphs, Symmetric graphs, s-Regular graphs, Regular coverings.

1. Introduction

Throughout this paper graphs are assumed to be finite, simple, undirected and connected. For group theoretic concepts and notation not defined here, we refer the reader to [13, 24]. Given a positive integer n, we shall use the symbol to denote the ring of residues modulo as well as the cyclic group of order . For a graph X, we use V(X), E(X) and Aut(X) to we denote its set, edge set and automorphism group, respectively. For , , is the edge incident to and in X and is the set of vertices adjacent to in X. For a subgroup N of Aut(X), denote by XN the quotient graph of X corresponding to the orbits of N, that is, the graph having the orbits of N as vertices with two orbits adjacent in XN whenever there is an edge between those orbits in X. A graph is called a covering of a graph with projection :→ if there is a surjection : → such that | is a bijection for any vertex and 1. A covering of with a projection is said to be regular (or k-covering) if there is a semiregular subgroup K of the automorphism group Aut() such that graph is isomorphic to the quotient graph K, say by h, and the quotient map → K is the composition h of and h. If is connected K becomes the covering transformation group. The fibre of an edge or vertex is its preimage under . An automorphism of is said to be fibre-preserving if it maps a fibre to a fibre, while every covering transformation maps a fibre onto itself. All of fibre-preserving automorphism from a group called the fibre-preserving group. Let G be a finite group and S a subset of G such that 1S and S = S-1 = {s−1|s S}. The Cayley graph Cay (G, S) on G with respect to S is defined to have

1 Prof., Department of Mathematics, Faculty of Science, South Tehran branch, Islamic Azad University, Tehran, Iran, e-mail: [email protected]. 2 Department of Mathematics, Iran University of Science and Technology, Iran 220 Mehdi Alaeiyan, M. K. Hosseinipoor

vertex set G and edge set {gh| g,h G,gh−1 S}. A Cayley graph Cay (G, S) is connected if and only if S generates G. It is well known that Aut (Cay (G, S)) contains the right regular representation R(G) of G, the acting group of G by right multiplication, which is regular on vertices. A Cayley graph Cay (G, S) is said to be normal if R(G) is normal in Aut(Cay(G,S)). A graph is isomorphic to a Cayley graph on G if and only if Aut() has a subgroup isomorphic to G, acting regularly on vertices (see [4, Lemma 16.3]). Let s be a positive integer. An s-arc in a graph X is an ordered (s+1)-tuple ,,…,, of vertices of X such that 1 is adjacent to for 1 ≤ i ≤ s and 1 1 for 1 ≤ i ≤ s − 1. For a graph X and a subgroup G of Aut(X), X is said to be G-vertex-transitive, G-edge-transitive and G-s-arc-transitive if G acts transitively on the sets of vertices, edges and s-arcs of X respectively. It is easily seen that a graph X which is G-edge but not G-vertex-transitive is necessarily bipartite, with the two parts of bipartition coinciding with the orbits of G. In particular, if X is a regular graph, then these two parts have equal cardinalities, and such a graph is then referred to as being G-semisymmetric. In the case where G =Aut(X) the symbol G may be omitted from the definitions above, so that a graph X is called vertex-transitive, edge-transitive, s-arc-transitive and semisymmetric if it is Aut(X)-vertex-transitive, Aut(X)-edge-transitive, Aut(X)-s- arc-transitive and Aut(X)-semisymmetric, respectively. In particular 1-arc- transitive means arc-transitive or symmetric. A symmetric graph X is said to be s- regular if Aut(X) acts regularly on the set of s-arcs in X. Tutte [21, 23] showed that every cubic symmetric graph is s-regular for some 1≤ s ≤5. Tutte [22], proved that a vertex- and edge-transitive graph with odd valency must be symmetric. Thus a cubic edge-transitive graph is either symmetric or semisymmetric. The classification of cubic symmetric or semisymmetric graphs of different order is given in many papers. For example, the cubic symmetric graphs of order 4p, 4,6,6,10,10,16 and 16p2 were classified in [12,11,19,3] and the cubic semisymmetric graphs of order 2p3,6p2,6p3,8p and 8p2 were classified in [18,16,9,2,1]. In this paper, we obtain a classification of cubic edge-transitive graphs of order 18p. In order to state the main Theorem 1.1 we first introduce a family of cubic graphs. Let p be a prime such that p ≡ 1 (mod 3), and let k be an element of order 3 in . Set V (K3,3) = {a,b,c,x,y,z} to be the vertex set of the complete K3,3 with partite sets {a,b,c} and {x,y,z}. The graph CF18p is defined to have vertex set V (CF18p) = V (K3,3) × 3p and edge set

E(CF18p) = {(a, i)(x, i), (a, i)(y, i), (a, i)(z, i), (b, i)(y, i), (b, i)(x, i + k + 1), (b, i)(z, i+1), (c, i)(x, i−1), (c, i)(y, i−k−1), (c, i)(z, i)|i 3p}. A classification of cubic edge-transitive graphs of order 18p 221

It will be shown in the Lemma 2.7 that the graph CF18p is independent of the choice of k and hence unique for given order. The following is the main result of this paper. Theorem 1.1 Let X be a connected cubic edge-transitive graph of order 18p, where p is a prime. Then X is either semisymmetric or s-regular for some s = 1, 2 or 5. Furthermore, (1) X is semisymmetric if and only if X is isomorphic to one of the graphs S54 and S126; (2) X is 1-regular if and only if X is isomorphic to the graph CF18p, where p ≡ 1 (mod 3); (3) X is 2-regular if and only if X is isomorphic to the graph F54; (4) X is 5-regular if and only if X is isomorphic to one of the graphs F90 and F234B. 2. Preliminaries Proposition 2.1 [18, Proposition 2.6] Let X be a G-semisymmetric graph for some subgroup G of Aut(X). Then either X ≡ K3,3 or G acts faithfully on each of bipartition sets of X. Proposition 2.2 [18, Proposition 2.3] Let X be a connected bipartite graph admitting an abelian subgroup G ≤ Aut(X) acting regularly on each of the bipartition sets. Then X is vertex-transitive. The next Proposition is a special case of [16, Lemma 3.2]. Proposition 2.3 Let X be a connected G-semisymmetric with bipartition sets L(X) and R(X) and let N be a normal subgroup of G. If N is intransitive on bipartition sets, then N acts semiregularly on both L(X) and R(X), and X is an N-covering of a G/N-semisymmetric graph. Proposition 2.4 [15. Theorem 9] Let X be a connected symmetric graph of prime valency and G an s-arc-transitive subgroup of Aut(X) for some s ≥ 1. If a normal subgroup N of G has more than two orbits, then it is semiregular and G/N is an s-arc-transitive subgroup of Aut(XN), where XN is the quotient graph of X corresponding to the orbits of N. Furthermore, X is a regular covering of XN with the covering transformation group N. Let X = Cay (G, S) be a Cayley graph on a group G with respect to S. Set A :=Aut(X) and Aut(G,S) := {α Aut(G) | Sα = S}. Then we have: Proposition 2.5 [25, Proposition 1.5] The Cayley graph X = Cay (G,S) is normal if and only if A1 =Aut(G,S), where A1 is the stabilizer of the vertex 1 V (X) = G in A. Let p be a prime. It is easy to see that the equation x2 + x + 1 = 0 (1) has no solution in the ring 3 for p = 3. The following result determines the solutions of Eq. (1) in 3 for p ≠ 3. 222 Mehdi Alaeiyan, M. K. Hosseinipoor

Lemma 2.6 Let p ≠ 3 be a prime. Then there exists k 3 solving Eq. (1) if and only if k is an element of order 3 in . 2 Proof. Suppose first that k 3p such that k + k + 1 = 0. Then k ≠ 1 and since k3 − 1 = (k − 1)(k2 + k + 1) = 0, it follows that k is an element of order 3 in . Conversely, suppose that k is an element of order 3 in . Then k ≠ 1 and k3 = 1. It follows that (k − 1)(k2 + k + 1) = 0. If k − 1 is divisible by 3, then k2 + k + 1 is also divisible by 3. Thus, in order to prove k2 + k + 1 = 0, it suffices to show k − 1 is coprime with p. Assume that (k − 1, p) ≠ 1. Then k ≡ 1 (mod p). Let k = tp + 1. Then k3 = t3p3 + 1. Since k3 = 1 and p = 36, we have t ≡ 0 (mod 3). Hence k = 1, a contradiction. This completed the proof of lemma. Let p be a prime such that 1 3. , by 2 Lemma 2.6, there are exactly two elements of order 3, say k and k in solving Eq. (1). Denote by V (K3,3) = {a,b,c,x,y,z} the vertex set of K3,3 as before. The graphs

CF18p and 18 are defined to have the same vertex set V (CF18p) = V

(18) = V (K3,3) × 3p and edge sets E(CF18p) = {(a,i)(x,i), (a,i)(y,i), (a,i)(z,i), (b,i)(y,i), (b,i)(x,i + k + 1), (b,i)(z,i+1), (c,i)((x,i−1), (c,i)(y,i−k−1), (c,i)(z,i)|i 3p}, 2 E(CF18p) = {(a,i)(x,i), (a,i)(y,i), (a,i)(z,i), (b,i)(y,i), (b,i)(x,i + k + 1), 2 (b,i)(z,i+1), (c,i)((x,i−1), (c,i)(y,i−k −1), (c,i)(z,i)|i 3p},

2 respectively. The graph 18 is obtained by replacing k with k in each edge of

CF18p. It is easy to see that CF18p and 18 are cubic and bipartite.

Lemma 2.7 The graphs CF18p and 18 are isomorphic. Proof. Let p be a prime such that p 1(mod 3) and k an element of order 3 in . To show CF18p we define a map α from V (CF18p) to V (18) by (a,i) Æ (a,ki), (b,i) Æ (c,ki), (c,i) Æ (b,ki), (x,i) Æ (x,ki), (y,i) Æ (z,ki), (z,i) Æ (y,ki), where i 3. Clearly,

18, y, , x, 1,z,1, 18, 18, =, 1, , 1,,

By Lemma 2.6, k2 + k + 1 = 0. With use of this property, one can show that α α [NCF18p((b,i))] = ((b,i) ), Similarly, α α [NCF18p((u,i))] = ((u,i) ), A classification of cubic edge-transitive graphs of order 18p 223

for u = a, c. It follows that α is an isomorphism from CF18 to 18, because the graphs are bipartite. In view of [17, Corollary 2.2], [10, Theorem 1.1], Lemma 2.6 and Lemma 2.7 imply the following.

Lemma 2.8 Let X be a connected cubic 3-covering of K3,3 whose fibre- preserving group acts edge-transitively on X. Then p ≡ 1(mod 3) and X is isomorphic to the 1-regular graph CF18p. 3. Proof of Theorem 1.1 Lemma 3.1 Let p be a prime. Then, with the exception of the graphs S54 (the Gray graph) and S126, every connected cubic edge-transitive graph of order 18p is symmetric. Proof. By [7], there are two cubic semisymmetric graphs of order 18p for p ≤ 41 which are the graphs S54 and S126. To prove the lemma, we only need to show that no connected cubic semisymmetric graph of order 18p exists for p ≥ 43. Suppose that there exists a connected cubic semisymmetric graph X of order 18p with p ≥ 43. Then X is bipartite. Denote by L(X) and R(X) the bipartition sets of X. Clearly, |L(X)| = |R(X)| = 9p. Set A:=Aut(X). By Proposition 2.1, A is faithful on each bipartition sets of X and by [18, Proposition 2.4], the vertex r r 3 stabilizer Av of v V (X) has order 2 .3, where r ≥ 0. Thus |A| = 2 .3 .p. By [14, pp. 12- 14], there is no simple {2,3, p}-group for p ≥ 43. Hence A is solvable. For any prime divisor q of |A|, denote by Oq(A) the largest normal q- subgroup of A. Then by Proposition 2.3, Oq(A) is semiregular on L(X) and R(X) and the quotient graph XOq(A) is A/Oq(A)-semisymmetric. Thus O2(A) = 1 and |O3(A)| is a divisor of 9. By the solvability of A, either O3(A) ≠ 1 or Op(A) ≠ 1. If O3(A) ≠ 1 then |O3(A)| = 3 or 9. Let T/O3(A) be a minimal normal subgroup of A/O3(A). Since A is solvable, A/O3(A) is also solvable, implying T/O3(A) is elementary abelian. The quotient graph XO3(A) is A/O3(A)-semisymmetric. Thus by Proposition 2.3, O2(A/O3(A)) = 1 and since O3(A/O3(A)) = 1, we conclude that T/O3(A) is a p-group. Now |T| = 3p or 9p and since p ≥ 43, by Sylow Theorem, T has a normal Sylow p-subgroup, which is characteristic in T and hence normal in A because T A. Therefore Op(A) ≠ 1. Set P = Op(A). Since Op(A) ≠ 1, we have P p. Note that the quotient graph XP is A/P-semisymmetric. Thus XP is bipartite. Denote by L(XP) and R(XP) the bipartition sets of XP . Then |L(XP)| = |R(XP)| = 9. Set C = CA(P). Then P ≤ C. Suppose C = P. Then by [20, Theorem 1.6.13], A/P is isomorphic to a subgroup of Aut(P) p−1, which implies that A/P is abelian. Since XP is A/P semisymmetric, it follows by Proposition 2.1, and [24, Proposition 4.4] that A/P is regular on L(XP) (and also R(XP )). This implies that |A| = 9p, a contradiction. Thus P < C, that is, P is a proper subgroup of C. Let M/P be a minimal normal subgroup of A/P contained in C/P. Then M/P is an elementary abelian 3-group because by Proposition 2.3, O2(A/O3(A)) = 1. Let Q be a Sylow 3-subgroup of M. Then M = 224 Mehdi Alaeiyan, M. K. Hosseinipoor

PQ, implying M = P × Q because Q < C. Hence Q is characteristic in M and since M A, we have Q A. Thus Q ≤ O3(A), it follows that |Q| = 3 or 9 and Q is semiregular on L(XP ) and R(XP ). Let |Q| = 9. Then M is an abelian group of order 9p. Since the vertex stabilizer of each vertex of X has order 2r.3, it follows by the semiregularity of Q on L(X) and R(X) that M is regular on each of the bipartition sets of X. By Proposition 2.2, X is vertex-transitive, a contradiction. Thus Q 3, and hence M 3p. By Proposition 2.3, X is a 3p-covering of the bipartite graph K3,3 and since M A, the fibre-preserving group is the automorphism group A of X, so it is edge-transitive. But by Lemma 2.8, X is symmetric, which is a contradiction. Lemma 3.2 Let X be a connected cubic symmetric graph of order 18p, where p is a prime. Then X is 1-,2-, or 5-regular. Furthermore, (1) X is 1-regular if and only if X is isomorphic to the 1-regular graph CF18p, where p ≡ 1 (mod 3); (2) X is 2-regular if and only if X isomorphic to the graph F54; (3) X is 5-regular if and only if X is isomorphic to one of the graphs F90 and F234B. Proof. Let X be a cubic graph satisfying the assumptions and let A =Aut(X). Since X is symmetric by Tutte [23], X is s-regular for some s ≤ 5. Thus |A| = 2s.33.p. For each prime p = 7,19,31 or 37, by [6] and Lemma 2.8, there is only one connected cubic symmetric graph of order 18p, which is the 1-regular graph CF18p and for each prime p = 2,11,17,23,29 or 41, there is no connected cubic symmetric graph of order 18p. Similarly, for p = 3 or 5, there is only one connected cubic symmetric graph of order 18p, that is, the 2-regular graph F54 and the 5-regular graph F90, and for p = 13 there are two connected cubic symmetric graph of order 18 × 13 which are the 1-regular graph CF234 and the 5-regular graph F234B. Thus we may assume that p ≥ 43. By [14, pp. 12-14], A is solvable. Let q be a prime divisor of |A|. Then by Proposition 2.4, Oq(A) is semiregular on V (X) and the quotient graph XOq(A) is a cubic symmetric graph. 2 The semiregularity of Oq(A) implies that |Oq(A)| | 18p . If O2(A) = 16, then O2(A) 2 and hence XO2(A) has odd order and valency 3, a contradiction. Thus O2(A) = 1 and by the solvability of A, either O3(A) ≠ 1 or Op(A) ≠ 1. If O3(A) ≠ 1, then |O3(A)| = 3 or 9. Let T/O3(A) be a minimal normal subgroup of A/O3(A). Then T/O3(A) is elementary abelian because A is solvable. Since O3(A/O3(A)) = 1, T/O3(A) is a 2- or p-group. For the former by Proposition 2.4, the quotient graph XT would have be a cubic graph of odd order, a contradiction. Thus T/O3(A) is a p-group. Now |T| = 3p or 9p and since p ≥ 43, by Sylow Theorem, M has a normal Sylow p-subgroup, which is characteristic in T and hence normal in A. Therefore Op(A)≠1. A classification of cubic edge-transitive graphs of order 18p 225

Set P = Op(A). Since Op(A) ≠ 1, we have P p. By Proposition 2.4, the quotient graph XP is a cubic symmetric graph and A/P is an arc-transitive subgroup of Aut(XP ). Set C = CA(P). Clearly P ≤ C. Suppose that P = C. Then by [20, Theorem 1.6.13], A/P is isomorphic to a subgroup of Aut(P) p−1, which implies that A/P is abelian. Since A/P is transitive on V (XP ), it follows by [24, Proposition 4.4] that A/P is regular on V (XP). This implies that |A| = 18p, a contradiction. Thus P < C. Let M/P be a minimal normal subgroup of A/P contained in C/P. Then M/P is an elementary abelian 2- or 3-group. If M/P is a 2- group then by Proposition 2.4, the quotient graph XM is a cubic symmetric graph of odd order, a contradiction. Thus M/P is a 3-group. Let Q be a Sylow 3- subgroup of M. Then M = PQ, implying M = P × Q because Q < C. Hence Q is characteristic in M and since M A, we have Q A. Thus Q ≤ O3(A) and hence |Q| = 3 or 9. Note that Q is isomorphic to the elementary abelian group M/P. Then or . Let . Then the quotient graph XQ is a cubic symmetric graph of order 2p and A/Q is an arc-transitive subgroup of Aut(XQ). Since p ≥ 43, by [5] and [8, Lemma 3.4], XQ is a normal cubic 1-regular Cayley graph on dihedral group D2p. Thus A/Q =Aut(XQ) and A has a normal subgroup G such that G/Q acts regularly on V (XQ). Consequently, G is regular on V (X) and hence X is a normal cubic 1-regular Cayley graph on G. Let X = Cay (G, S). Since X has valency 3, S contains at least one involution. By Proposition 2.5, Aut(G,S) is transitive on S, which implies that S consists of three involutions and by the connectivity of X, G can be generated by three involutions. Note that M/Q is a normal Sylow p- subgroup of A/Q. Then M/Q G/Q, implying M G. Since . and G/Q D2p, one can conclude that , which is impossible because in each case G cannot be generated by involutions. Thus Q 3, and so M 3p. By Proposition 2.4, X is a 3p-covering of the bipartite graph K3,3 and since M A, the symmetry of X means that the fibre- preserving group is arc-transitive and hence is edge-transitive. By Lemma 2.8, X is isomorphic to CF18p, where p ≡ 1 (mod 3). Proof of Theorem 1.1 It follows by Lemmas 3.1 and 3.2.

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